Question

(3-√21)+(2√3-5√21)-(4√3+4√21)
3-√21+2√3-5√21-4√3+4√21

Answer 1

Question: Can you solve √ (21-4√5+8√3-4√15) here first square root is over all the equation?

Solution:

[math]\sqrt{21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}}[/math]

The clue lies in the terms [math]\sqrt{5}, \sqrt{3}, \sqrt{15} [/math]Note: [math]\sqrt{15} = \sqrt{3} * \sqrt{5}[/math]

Also, there is a constant term. One can conclude that expression when simpiefied must be of the form [math](a + b\sqrt{3} + c\sqrt{5})[/math]

[math]\therefore[/math] [math]\sqrt{21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}} = (a + b\sqrt{3} + c\sqrt{5}) \tag {E01}[/math]

Squaring both sides:

[math]\implies 21 - 4\sqrt{5} + 8\sqrt{3} - 4\sqrt{15} = (a + b\sqrt{3} + c\sqrt{5})^2[/math]

[math]\implies 21 - 4\sqrt{5} + 8\sqrt{3} - 4\sqrt{15} = a^2 + 3b^2 + 5c^2 + 2ab\sqrt{3} + 2bc \sqrt{15} + 2ca \sqrt{5}[/math]

Comparing the like terms on both sides:

[math]\sqrt{15}[/math] term: [math]-4 = 2bc \implies bc = -2 \tag {E02}[/math]

[math]\sqrt{5}[/math] term: [math]-4 = 2ca \implies ca = -2 \tag {E03}[/math]

[math]\sqrt{3}[/math] term: [math]8 = 2ab \implies ab = 4 \tag {E04}[/math]

From [math](E02)[/math] and [math](E03): \dfrac{bc}{ca} = \dfrac{-2}{-2} \implies a = b[/math]

From [math](E04): ab = 4 \implies a^2 = 4 \implies a = b = \pm 2[/math]

From [math](E02): bc = -2 \implies c = \mp 1[/math]

[math]\therefore (E01): \boxed{\sqrt{21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}} = \pm (2 + 2\sqrt{3} - \sqrt{5})}[/math]

Validation:

[math](2 + 2\sqrt{3} - \sqrt{5})^2 = 2^2 + (2\sqrt{3})^2 + (\sqrt{5})^2 + (2\cdot 2 \cdot 2\sqrt{3}) - (2 \cdot 2\sqrt{3} \cdot \sqrt{5}) - (2\cdot 2 \cdot \sqrt{5})[/math]

[math]= 4 + 12 + 5 + 8\sqrt{3} - 4\sqrt{15}- 4\sqrt{5} [/math]

[math]= 21 + 8\sqrt{3} - 4\sqrt{15}- 4\sqrt{5} = [/math]Given expression. So, answer is good.